Posts Tagged ‘Systoles’

This is a little note about constructing a Riemannian 2-sphere which has longer shortest geodesic than the round 2-sphere of same area.

—– Background Story —–

So there has been this thing called ‘mathematical conversations’ at the IAS, which involves someone present a topic that’s elementary enough to be accessible to mathematicians in all fields and yet can be expanded in different directions and lead into interesting interdisciplinary discussions.

Nancy Hingston gave one of those conversations about simple geodesics on the two-sphere one night and I was (thanks to Maria Trnkova who dragged me in) able to attend.

So she talked about some fascinating history of proving the existence of closed geodesics and later simple closed geodesics on generic Riemannian two-spheres.

Something about this talk obviously touched my ‘systolic nerve’, so when the discussion session came up I asked whether we have bounds on ‘length of longest possible shortest closed geodesic on a sphere with unit area’. The question seem to have generated some interest in the audience and resulted in a back-and-forth discussion (some of which I had no clue what they were talking about). So the conclusion was at least nobody knows such a result on top of their head and perhaps optimum is obtained by the round sphere.

—– End of Story —-

A couple of post-docs caught me afterwards (Unfortunately I didn’t get their names down, if you happen to know who they are, tell me~) and suggested that suspending a smooth triangular region and smoothen the corners might have longer shortest geodesic than the round sphere:

The evidence being the fact that on the plane a rounded corner triangular contour has larger ‘width’ than the disc of same area. (note such thing can be made to have same width in all directions)

Well that’s pretty nice, so I went home and did a little high-school computations. The difficulty about the pillow is that the shortest geodesic isn’t necessarily the one that goes through the ‘tip’ and ‘mid-point of the base’, something else might be shorter. I have no idea how to argue that.

A suspicious short geodesic:

So I ended up going with something much simpler, namely gluing together two identical copies of the flat equilateral triangles. This can be made to a Riemannian metric by smoothing the edge and corners a little bit:

Okay, now the situation is super simple~ I want to prove that this ‘sphere’ (let’s call it from now on) has shortest geodesic longer than the round sphere ()!

Of course we suppose both and has area 1.

Claim: The shortest geodesic on has length (which is length of the one through the tip and mid-point of the opposite edge.)

Proof: The shortest closed geodesic passing through the corner is the one described above, since any other such geodesics must contain two symmetric segments from the corner to the bottom edge on the two triangles, those two segments alone is longer than the one orthogonal to the edge.

That middle one has length where

i.e.

The good thing about working with flat triangles is that now I know what the closed geodesics are~

First we observe any closed geodesic not passing through the corner is a periodical billiard path in the triangular table with even period.

So let’s ‘unfold’ the triangles on the plane. Such periodic orbits correspond to connecting two corresponding points on a pair of identified parallel edges and the segment between them intersecting an even number of tiles.

W.L.O.G we assume the first point in on edge . Since we are interested in orbits having shortest length, let’s take neighborhood of radius around our edge : (all edges with arrows are identified copies of )

There are only 6 parallel copies of in the neighborhood:

Note that no matter what point on we start with, the distance from to another copy of it on any of the six edges is EQUAL to . (easy to see since one can slide the segments to begin and end on vertices.)

Hence we conclude there are no shorter periodic billiard paths, i.e. the shortest closed geodesic on has length .

Note it’s curious that there are a huge amount of closed geodesics of that particular length, most of them are not even simple! However it seems that after we smoothen to a Riemannian metric, the non-simple ones all become a little longer than that simple one through the corner. I wonder if it’s possible that on a Riemannian sphere the shortest closed geodesic is a non-simple one.

Anyways, now let’s return to ~ So the surface area is hence the radius is

Any closed geodesics is a multiple of a great circle, hence the shortest geodesic has length , which is just slightly shorter than .

Now the natural question arises: if the round sphere is not optimum, then what is the optimum?

At this point I looked into the literature a little bit, turns out this problem is quite well-studied and there is a conjecture by Christopher Croke that the optimum is exactly . (Of course this optimum is achieved by our singular triangle metric hence after smoothing it would be .

There is even some recent progress made by Alex Nabutovsky and Regina Rotman from (our!) University of Toronto! See this and this. In particular, one of the things they proved was that the shortest geodesic on a unit area sphere cannot be longer than , which I believe is the best known bound to date. (i.e. there is still some room to .)

Random remark: The essential difference between this and the systolic questions is that the sphere is simply connected. So the usual starting point, namely ‘lift to universal cover’ for attacking systolic questions does not work. There is also the essential difference where, for example, the question I addressed above regarding whether the shortest geodesic is simple would not exist in systolic situation since we can always split the curve into two pieces and tighten them up, at least one would still be homotopically non-trivial. In conclusion since this question sees no topology but only the geometry of the metric, I find it interesting in its own way.

For a compact subset of , we define the k-codimensional width (or simply k-width) to be the smallest possible number where there exists a k-dimensional affine subspace s.t. all points of is no more than away from .

i.e.

where is the length of the orthogonal segment from to .

It’s easy to see that, for any ,

.

At the first glance it may seems that . However it is not the case since for example the equilateral triangle of side length in has diameter but 0-width . In fact, by a theorem of Jung, this is indeed the optimum case, i.e. we have:

At this point one might wonder (at least I did), if we want to invent a notion that captures the ‘diameter’ after we ‘forget the longest k-dimensions’, a more direct way seem to be taking the smallest possible number where there is an orthogonal projection of onto a dimensional subspace where any point has pre-image with diameter .

i.e.

Now we easily have .

However, the disadvantage of this notion is, for example, there is no reason for a semicircle arc to have 1-width 0 but a three-quarters circular arc having positive 1-width.

Since we are measuring how far is the set from being linear, taking convex hull should not make the set ‘wider’ , unlike is not invariant under taking convex hulls. Note that for convex sets we do have

iff is contained in a -plane.

We now generalize this notion to general metric spaces:

Definition: The Uryson k-width of a compact metric space is the smallest number where there exists dimensional topological space and a continuous map where any point has pre-image with diameter .

i.e.

Note: Here dimension is the usual covering dimension for topological spaces: i.e. a topological space is dimensional if any finite cover of has a finite refinement s.t. no point of is contained in more than sets in the cover and is the smallest number with this property.

For compact subsets of with induced metric, we obviously we have since the pair is clearly among the pairs we are minimizing over.

Speaking of topological dimensions, one of the classical results is the following:

Lebesgue’s lemma: Let be the solid n-dimensional cube, then for any topological space with and any continuous map , we have image of at least one pair of opposite -faces intersect.

Since the conclusion is purely topological, this applies equally well to rectangles. i.e. for , , we have ; furthermore, for all .

(If the later statement does not hold, we write as , being the product of the first coordinates. Now ).

In light of the earlier post about minimax inequality, we should note that if we restrict to be a homeomorphic copy of then the notion is the same as the minimax length of fibres. In particular as proved in the post the minimax length of the unit disc to is 2.

Exercise: Check that for the unit -disk, , i.e. the optimum is obtained by contracting the disc onto a triod.

Hence it can indeed be strictly smaller than merely taking as the targeting space, even for simply connected sets. This gives a better measurement of ‘width’ in the sense that, for example, the neighborhood of a tree will have about .

The goal for most of the posts in this blog has been to take out some very simple parts of certain papers/subjects and “blow them up” to a point where anybody (myself included) can understand. Ideally the simple parts should give some inspirations and ideas towards the more general subject. This one is on the same vein. This one is based on parts of professor Guth’s minimax paper.

In an earlier post, we talked about the extremal length where one is able to bound the “largest possible minimum length” (i.e. the “maximum minimum length“) of a family of rectifiable curves under conformal transformation. When combined with the uniformization theorem in for surfaces, this becomes a powerful tool for understanding arbitrary Riemannian metrics (and for conformal classes of metrics in higher dimensions).

However, in ‘real life’ we often find what we really want to bound is, instead, the “minimum maximum length” of a family of curves, for example:

Question: Let be the unit disc. Given any family of arcs with endpoints on and foliates , then how short can the logest arc in possibly be?

In other words, let be the collection of all possible such foliations as above, what is

?

After playing around a little bit with those foliations, we should expect one of the fibres to be at least as long as the diameter ( i.e. no foliation has smaller maximum length leaf than foliating by straight lines ). Hence we should have

.

This is indeed easy to prove:

Proof: Consider the map where , switches the end-points of each arc in . It is easy to check that is a continuous, orientation reversing homeomorphism of the circle (conjugate to a reflection). Let be its fixed points, be the two arcs in connecting to .

Let

be the antipodal map on .

Suppose then one of is longer than , say it’s .

Then we have

.

Hence has a fixed point in , i.e. .

There is a fibre in with endpoints , the fibre must have length

.

The remaining case is trivial: if then both and gets mapped into themselves orientation-reversingly, hence fixed points still exists.

Establishes the claim.

Instead of the disc, we may look at circles that sweep out the sphere (hence to avoid the end-point complications):

Theorem: Any one-parameter family of circles that foliates (except two points) must have the largest circle being longer than the equator.

This is merely applying the same argument, i.e. one of the circles needs to contain a pair of antipodal points hence must be longer than the equator.

In order for easier generalization to higher dimensions, with slight modifications, this can be formulated as:

Theorem: For any having non-zero degree, there is where is larger than the equator.

Hence in higher dimensions we can try to prove the same statement for largest image of a lever -sphere under . However before we do that I would like to highlight some intergal geometry machineries that are new to me but seemingly constantly used in proving those kinds of estimates. We shall get some idea of the method by showing:

Theorem: Let be equipped with the round metric. be a ‘flat’ -dimensional plane. Then any -chain in the same dimensional homology class as must have volume at least as large as .

Proof: Let be the set of all -planes in (i.e. the Grassmannian).

There is a standard way to associate a measure on :

Let be the Haar measure on , fix some .

Since acts on , for open set , we set

.

–The measure of a collection of planes is the measure of linear transformations that takes the given plane to an element of the set.

Now we are able to integrate over all -planes!

For almost all , since is -plane, we have . ( not only when they are ‘parallel’ )

Since in , for almost all , intersects at least as much as does. We conclude that for almost all .

Fact: There exists constant such that for any -chain ,

.

The fact is obtained by diving the chain into fine cubes, observe that both volume and expectation are additive and translation invariant. Therefore we only need to show this for infinitesimal cubes (or balls) near . We won’t work out the details here.

Hence in our case, since for almost all we have , the expectation .

We therefore deduce

.

Establishes the theorem.

Remark: I found this intergal geometry method used here being very handy: in the old days I always try to give lower bounds on volume of stuff by intersecting it with planes and then pretend the ‘stuff’ were orthogonal to the plane, which is the worst case in terms of having small volume. An example of such bound can be found in the knot distorsion post where in order to lower bound the length we look at its intersection number with a family of parallel planes and then integrate the intersection.

This is like looking from one particular direction and record how many times did a curve go through each height, of course one would never get the exact length if we know the curve already. What if we are allowed to look from all directions?

I always wondered if we know the intersection number with not only a set of parallel planes but planes in all directions, then are there anything we can do to better bound the volume? Here I found the perfect answer to my question: by integrating over the Grassmannian, we are able to get the exact volume from how much it intersect each plane!

We get some systolic estimates as direct corollaries of the above theorem, for example:

Corollary: where carries the round metric with total volume .

Back to our minimax problems, we state the higher dimensional version:

Wish: For any map where carries the standard round metric, there exists some with

where is the -dimensional equator.

But what we have is that there is a (small) positive constant s.t. implies

(shown by an inductive application of the isomperimetric inequality on , which is obtained from applying intergal geometry methods)

Recently I came across a paper by John Pardon – a senior undergrad here at Princeton; in which he answered a question by Gromov regarding “knot distortion”. I found the paper being pretty cool, hence I wish to highlight the ideas here and perhaps give a more pictorial exposition.

This version is a bit different from one in the paper and is the improved version he had after taking some suggestions from professor Gabai. (and the bound was improved to a linear one)

Definition: Given a rectifiable Jordan curve , the distortion of is defined as

.

i.e. the maximum ratio between distance on the curve and the distance after embedding. Indeed one should think of this as measuring how much the embedding ‘distort’ the metric.

Given knot , define the distortion of to be the infimum of distortion over all possible embedding of :

It was (somewhat surprisingly) an open problem whether there exists knots with arbitrarily large distortion.

Question: (Gromov ’83) Does there exist a sequence of knots where ?

Now comes the main result in the paper: (In fact he proved a more general version with knots on genus surfaces, for simplicity of notation I would focus only on torus knots)

Theorem: (Pardon) For the torus knot , we have

.

To prove this, let’s make a few observations first:

First, fix a standard embedding of in (say the surface obtained by rotating the unit circle centered at around the -axis:

and we shall consider the knot that evenly warps around the standard torus the ‘standard knot’ (here’s what the ‘standard knot looks like:

By definition, an ’embedding of the knot’, is a homeomorphism that carries the standard to the ‘distorted knot’. Hence the knot will lie on the image of the torus (perhaps badly distorted):

For the rest of the post, we denote by and by , w.l.o.g. we also suppose .

Definition: A set is inessential if it contains no homotopically non-trivial loop on .

Some important facts:

Fact 1: Any homotopically non-trivial loop on that bounds a disc disjoint from intersects at least times. (hence the same holds for the embedded copy ).

As an example, here’s what happens to the two generators of (they have at least and intersections with respectively:

From there we should expect all loops to have at least that many intersections.

Fact 2: For any curve and any cylinder set where is in the -plane, let we have:

i.e. The length of a curve in the cylinder set is at least the integral over -axis of the intersection number with the level-discs.

This is merely saying the curve is longer than its ‘vertical variation’:

Similarly, by considering variation in the radial direction, we also have

Proof of the theorem

Now suppose , we find an embedding with .

For any point , let

is inessential

i.e. one should consider as the smallest radius around so that the whole ‘genus’ of lies in .

It’s easy to see that is a positive Lipschitz function on that blows up at infinity. Hence the minimum value is achieved. Pick where is minimized.

Rescale the whole so that is at the origin and .

Since (and note distortion is invariant under scaling), we have

Hence by fact 2,

i.e. There exists where the intersection number is less or equal to the average. i.e.

We will drive a contradiction by showing there exists with .

Let , since

By fact 2, there exists s.t. . As in the pervious post, we call a ‘neck’ and the solid upper and lower ‘hemispheres’ separated by the neck are .

Claim: One of is inessential.

Proof: We now construct a ‘cutting homotopy’ of the sphere :

i.e. for each is a sphere; at it splits to two spheres. (the space between the upper and lower halves is only there for easier visualization)

Note that during the whole process the intersection number is monotonically increasing. Since , it increases no more than .

Observe that under such ‘cutting homotopy’, is inessential then is also inessential. (to ‘cut through the genus’ requires at least many intersections at some stage of the cutting process, but we have less than many interesections)

Since is disconnected, the ‘genus’ can only lie in one of the spheres, we have one of is inessential. Establishes the claim.

We now apply the process again to the ‘essential’ hemisphere to find a neck in the direction, i.e.cutting the hemisphere in half in direction, then the -direction:

The last cutting homotopy has at most many intersections, hence has inessential complement.

Hence at the end we have an approximate ball with each side having length at most , this shape certainly lies inside some ball of radius .

Let the center of the -ball be . Since the complement of the ball intersects in an inessential set, we have is inessential. i.e.

There has been a couple of interesting talks recently here at Princeton. Somehow the term ‘extremal length’ came up in all of them. Due to my vast ignorance, I knew nothing about this before, but it sounded cool (and even somewhat systolic); hence I looked a little bit into that and would like to say a few words about it here.

Let be a simply connected Jordan domain in . is a conformal factor on . Recall from my last post, is a Lebesgue measurable function inducing a metric on where

and for any ( is an interval) with , we have the length of :

.

Call this metric on and denote metric space .

Given any set of rectifiable curves in (possibly with endpoints on ), each comes with a unit speed parametrization. Consider the “-width” of the set :

.

Let be the set of conformal factors with norm (i.e. having the total volume of normalized to ).

Definition: The extremal length of is given by

Remark: In fact I think it would be more natural to just use instead of since it’s called a “length”…but since the standard notion is to sup over all , not necessarily normalized, and having the -width squared divide by the volume of , I can’t use conflicting notation. One should note that in our case it’s just the square of sup of width.

Definition:The metric where this extremal is achieved is called an extremal metric for the family .

The most important fact about extremal length (also what makes it an interesting quantity to study) is that it’s a conformal invariant:

Theorem: Given bi-holomorphic, then for any set of normalized curves in , we can define after renormalizing curves in we have:

Sketch of a proof: (For simplicity we assume all curves in are rectifiable, which is not always the case i.e. for bad maps the length might blow up when the curve approach this case should be treated with more care)

This is indeed not hard to see, first we note that for any we can define by having

It’s easy to see that (merely change of variables).

In the same way, for any rectifiable curve.

Hence we have

.

On the other hand, we know that is a bijection from to , deducing

Establishes the claim.

One might wonder how on earth should this be applied, i.e. what kind of are useful to consider. Here we emphasis on the simple case where is a rectangle (Of course I would first look at this case because of the unresolved issues from the last post :-P ):

Theorem: Let , be the set of all curves starting at a point in the left edge , ending on with finite length. Then and the Euclidean metric is an extremal metric.

Sketch of the proof: It suffice to show that any metric with has at least one horizontal line segment with . (Because if so, and we know for the Euclidean length)

The average length of over is

By Cauchy-Schwartz this is less than

Since the shortest curve cannot be longer than the average curve, we have .

Corollary: Rectangles with different eccentricity are not conformally equivalent (i.e. one cannot find a bi-homomorphic map between them sending each edge to an edge).

Remark: I was not aware of this a few days ago and somehow had the silly thought that there are conformal maps between any pair of rectangles while discussing with Guangbo >.< then tried to see what would those maps look like and was of course not able to do so. (there are obviously Riemann maps between the rectangles, but they don't send conners to conners, i.e. can't be extended to a conformal map on the closed rectangle).

An add-on: While I came across a paper of Odes Schramm, applying the techniques of extremal length, the following theorem seemed really cool.

Let be a finite planar graph with vertex set and edges . For each vertex we assign a simply connected domain .

Theorem: We can scale and translate each to so that form a packing (i.e. are disjoint) and the contact graph of is . (i.e. iff .

Note: This is vastly stronger than producing a circle packing with prescribed structure.

About a year ago, I came up with an simple argument for the following simple theorem that appeared in a paper of professor Guth’s:

Theorem: If is an open set in the plane with area , then there is a continuous function from to the reals, so that each level set of has length at most .

Recently a question of somewhat similar spirit came up in a talk of his:

Question: Let be a Riemannian metric on the torus with total volume , does there always exist a function s.t. each level set of has length at most ?

I have some rough thoughts about how might a similar argument on the torus look like, hence I guess it would be a good idea to review and (somewhat carefully) write down the original argument. Since our final goal now is to see how things work on a torus (or other manifolds), here I would only present the less tedious version where is bounded and all boundary components of are smooth Jordan curves. Here it goes:

Proof: Note that if a projection of in any direction has length (one-dimensional measure) , then by taking to be the projection in the orthogonal direction, all level sets are straight with length (see image below).

Hence we can assume any -dimensional projection of has length . A typically bad set would ‘span’ a long range in all directions with small area, it can contain ‘holes’ and being not connected:

Project onto and -axis, by translating , we assume . Look at the measure set in the middle of (i.e. a measure 1 set with the property )

By Fubini, since the volume is at most , there must be a point with :

Since the boundary of is smooth, we may find a very small neighborhood where for each . (we will call this pink region a ‘neck’ of the set for it has small width and is roughly in the middle)

Now we define a that straches the neck to fit in a long thin tube (note that in general may not be connected, but everything is still well-defined and the argument does go through.) and then bend the neck to make the top chunk vertically disjoint from the bottom chunk.

We can take so that sends the vertical foliation of to the following foliation in (note that here we drew the neck wider for easier viewing, in fact the horizontal lines are VERY dense in the neck).

If the -projection of the top or bottom chunk is larger than , we repeat the above process t the chunks. i.e. Finding a neck in the middle measure set in the chunk, starch the neck and shift the top chunk, this process is guaranteed to terminate in at most steps. The final sends to something like:

Where each chunk has -width between and .

Define .

Claim: For any .

The vertical line intersects in at most one chunk and two necks, taking of the intersection, this is a PL curve with one vertical segment and two horizontal segment in :

The total length of is less than (length of on the vertical segment) (length of on each horizontal segment). Pick both less than , we conclude .

Establishes the theorem.

Remark:More generally,any open set of volume has such function with fibers having length . T he argument generalizes by looking at the middle set length set of each chunk.

Moving to the torus

Now let’s look at the problem on , by the uniformization theorem we have a flat torus where is a lattice, and a function s.t. is isometric to . is the flat metric. Hence we only need to find a map on with short fibers.

Note that

and the length of the curve from to in is

.

Consider as the parallelogram given by with sides identified. w.l.o.g. assume one side is parallel to the -axis. Let be a linear transformation preserving the horizontal foliation and sends the parallelogram to a rectangle.

Let be a piece-wise isometry that “folds” the rectangle:

(note that is four-to-one except for on the edges and the two medians)
Since all corresponding edges are identified, $lates F$ is continuous not only on the rectangle but on the rectangular torus.

Now we consider , pre-image of typical horizontal and vertical lines in the small rectangle are union of two parallel loops:

Note that vertical loops might be very long in the flat due to the shear while the horizontal is always the width.

Almost a year ago, I said here that I would write a sequence of posts on some simple facts and observations related to the systolic inequality but got distracted and didn’t manage to do much of that…

I was reminded last week as I heard professor Guth’s talk on systoles for the 4th time (Yes, the same talk! –in Toronto, Northwestern, India and here at the IAS). It’s interesting that I’m often thinking about different things each time I hear the same talk. This one is about the generalized Geroch conjecture.

Geroch conjecture: (the -torus) does not admit a metric of positive scalar curvature.

where is the Euclidian volume and is a positive constant only depending on the dimension .

Note that since our manifold does not have any cone points,

must vanish. Further more, the Riemannian structure on forces the term to vanish.

Since for this context we only care about is whether the scalar curvature is larger or smaller than , we can be even more simple-minded: has positive scalar curvature at all small enough balls around has smaller volume than their Euclidean cousins (with a difference of order propositional to ). In light of this definition, we have:

Restatement of the Geroch conjecture: For all on , there exists some point s.t. .

This is to say, small enough balls around some point $p$ are not small enough for it to have positive scalar curvature. What if instead we look at balls of a fixed radius instead of those infinitesimal balls? This naturally leads to

Generalized Geroch conjecture: For any , for all , there exists s.t. .

(For those larger than the injectivity radius, we lift to its universal cover so that all homotopically non-trival loops are ‘unfolded’)

Let’s take a look at the -torus to get a feel of the conjecture:

The flat torus, of course, has -scalar curvature at all points.

For the regular rotational torus, we take the ball around the saddle point of the gradient flow, the ball look like a saddle, as shown below.

To see that this has area larger than the analogous Euclidean ball, we can cut it along radial rays into thin triangles, each triangle can be ‘almost flattened’ to a Euclidean triangle, but we have a more triangles than in the Euclidean case.

What if we try to make the surface spherical for most of the area and having those negative scalar curvature points taking up a very small potion. One of my first attempts would be to connect a few spheres with cylinders:

We have a few parameters here: the number of balls , the width of the connecting cylinders , the length of the connecting cylinders and the radius of each sphere .

If cylinders are too long (longer than ), then we can just take the ball in the middle of the cylinder, the volume when lifted to universal cover would be equal to Euclidean.

If the width of cylinders are much smaller than , then the ball around a point in the gluing line would have volume almost a full spherical ball plus a half Euclidean ball, which would obviously be larger than a full Euclidean ball.

Hence the more interesting case is to have very short, wide tubes and as a consequence, have many balls forming a loop. In this case, the ‘worst’ ball would be centered at the middle of the tube, it intersects the two spheres connected by the tube in something a bit larger than a spherical half-ball.

I haven’t figured out an estimate yet. i.e. can the advantage taken from the fact that spherical ball are smaller than Euclidean balls cancel out the ‘a bit larger than half’? I think that would be interesting to work out.

Finally, let’s say what does this has to do with systoles:

Theorem: Generalized Geroch conjecture (which is the systolic inequality with a constant better than what we have so far)

Proof: Suppose not,

Let , by the generalized Geroch conjecture we have some larger than the Euclidean ball. i.e.

Since the systole is at least , hence cannot contain any homotopically non-trival loop i.e. it does not “warp around” and get unfolded when passing to the universal cover. Hence volume of a ball with radius cannot be larger than the volume of the whole manifold. Contradiction

Starting last summer with professor Guth, I’ve been interested in the systolic inequality for Riemannian manifolds. As a starting point of a sequence of short posts I plan to write on little observations I had related to the subject, here I’ll talk about the baby case where we find the lower bound of the systole on the -torus in terms of the area of the torus.

Given a Riemannian manifold where is the Riemannian metric.

Definition: The systole of is the length of smallest homotopically nontrivial loop in .

We are interested in bounding the systole in terms of the -th root of the volume of the manifold ( where is the dimension of ).

Note that the systole is only defined when our manifold has non-trivial fundamental group. I wish to remark that for the case of n-torus, having an inequality of the form is intuitive as we can see in the case of an embedded -torus in , we may deform the metric (hence the embedding) to make a non-contactable loop as small as we want while keep the volume constant, however when we attempt to make the smallest such loop large when not changing the volume, we can see that we will run into trouble. Hence it’s expected that there is an upper bound for the length of the smallest loop.

Since if only one loop in some homotopy class achieves that minimal length, we should be able to enlarge it and contract some other loops in that class to enlarge the systole and keep the volume constant. Hence it’s tempting to assume that all loops in the same class are of the same length. In the -torus case, such thing is the flat torus. Since any flat torus has systole proportional to , we have reasons to expect the optimal case fall inside this family. i.e.

.

This is indeed the case. The result was given in an early unpublished result by Loewner.

Let’s first optimize in the class of flat torus:

My first guess was that cannot be made less than i.e. the torus is the optimum case. However, this is not true. Let’s be more careful:

Since by scaling does not change ratio between and , we may normalize and let

Let be generators of the fundamental group of length of all geodesic loops in class are the side lengths of the parallelepiped i.e. and . W.L.O.G we suppose . has length and all geodesics in other classes are at least twice as long as one of the above three.

Hence the systole is maximized when those three are equal, we get . The systole in this case is and the volume is . Hence for any flat torus, we have

.

Theorem (Loewner): This bound holds for any metric on .

Proof: We will show this by reducing the case to flat metric.

induced an almost complex structure on , on surfaces, any almost complex structure is integrable. Hence there exists and where is a Riemann surface.

By uniformization theorem, is the quotient of by a discrete lattice. i.e. is a flat torus .

By scaling of the torus, we may assume the volume of the manifold is i.e.

Any nontrivial homotopy class of loops on can be represented by a straight loop on the flat torus. The length of such a loop in is merely integration of along the segment.

Here we have a family of loops in the homotopy class that is straight, by taking a segment of appropriate length orthogonal to the loops, we have the one-parameter family of parallel loops foliate the torus. Hence integrating over the segment of the length of the loops gives us the total volume of the torus. By Fubini, we have at least one loop is longer than volume of the torus over length of the segment we integrated on, which is the length of the straight loop in the flat torus.

Therefore the systole of is smaller than the minimum length of straight loops which is smaller than that of the flat torus. While the volume are the same. Hence it suffice to optimize the ratio in the class of flat tori. Establishes the theorem.

This is a note on Mikhail Katz’s paper (1995) in which he constructed a sequence of Riemannian metrics on s.t. for . Where denotes the -systole which is the infimum of volumes of -dimensional integer cycles representing non-trivial homology classes. To find out more about systoles, here’s a nice 60-second introduction by Katz.

We are interested in whether there is a uniform lower bound for for being equipped with any Riemann metric. For , it is known that . Hence the construction gave counterexamples for all . An counterexample for is constructed later using different techniques.

The construction breaks into three parts:

1) Construction a sequence of metrics on s.t. approaches as .

2) Choose an appropriate metric on s.t. equipped with the product metric satisfy the property

3) By surgery on to obtain a sequence of metrics on , denote the resulting manifolds by , having the property that

The first two parts are done in previous notes (which are not published on this blog). Here I will talk about how is part 3) done given that we have constructed manifolds as in part 2).

Let equipped with metric as constructed in 1), be as constructed in 2).

Standard surgery: Let and let . . The resulting manifold from standard surgery along in is defined to be which is homeomorphic to .

We perform the standard surgery on the component of , denote the resulting manifold by . Hence equipped with some metric.

Note that the metric depends on the surgery and so far we have only specified the surgery in the topological sense. Now we are going to construct the surgery taking the metric into account.

First we pick to be a small ball of radius , call it . Pick that fills to be a cylinder of length for some large with a cap on the top. i.e. and . Hence the standard surgery can be performed with and . The resulting manifold is homeomorphic to and has a metric on it that depends on and .

Let i.e. the part that’s glued in during the surgery, call it the ‘handle’.

The following properties hold:
i) For any fixed , for sufficiently small,

Since implies can be made small by taking small.

ii) The projection of to its factor is distance-decreasing.

iii) If we remove the the cap part from (infact from ), then the remaining part admits a distance-decreasing retraction to .

i.e. project the long cylinder onto its base on which is .

iv) Both ii) and iii) remain true if we fill in the last component of i.e. replace it with and get a -dimensional polyhedron .

Since all we did in ii) and iii) is to project along the first and third component simultaneously or to project only the first component, filling in the third component does not effect the distance decreasing in both cases.

We wish to choose an appropriate sequence of and so that .

In the next part we first fix any and so that property i) from above holds and write for .

We are first going to bound all cycles with a nonzero component and then consider the special case when the cycle is some power of and this will cover all possible non-trivial cycles.

Claim 1:-cycle belonging to a class with nonzero -component, we have .

Note that since and by part 2), and by property i), . Let , hence . Therefore the bound in claim 1 would imply which is what we wanted.

Proof:
a) If does not intersect

In this case the cycle can be “pushed off” the handle to lie in without increasing the volume. i.e. we apply the retraction from proposition iii).

b) If then by proposition ii), projects to its $S^n$ component by a distance-decreasing map and by construction in part 2).

Now suppose with .
Define s.t. .

Let , then by the coarea inequality, we have s.t. .

By our results in Gromov[83] and the previous paper of Larry Guth or Wenger’s paper, s.t. -cycle with , . Hence with . By picking , we have as .

Recall that ; by construction and .

Let ,

(1) If the cycle has non-trivial homology in , then by proposition iv), the analog of proposition iii) for implies we may retract to without decreasing its volume. Then apply case a) to the cycle after retraction we obtain .

Contradicting the assumption that .

(2) If has trivial homology in , then is a cycle with volume smaller than that’s contained entirely in . By case b), projects to its factor by a distance decreasing map, and . As above, , contradiction.